Question: The equation of a circle $C$ is $x^2+y^2-18x-14y+81 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2-18x) + (y^2-14y) = -81$ $(x^2-18x+81) + (y^2-14y+49) = -81 + 81 + 49$ $(x-9)^{2} + (y-7)^{2} = 49 = 7^2$ Thus, $(h, k) = (9, 7)$ and $r = 7$.